This is the submitted version of the article: Badr, Eslam; Bars Cortina, Francesc; Lorenzo García, Elisa. «On twists of smooth plane curves». Mathematics of Computation, Vol. 88, Issue 315 (March 2019), p. 421-438. DOI 10.1090/mcom/3317 This version is available at https://ddd.uab.cat/record/240667 under the terms of the license ON TWISTS OF SMOOTH PLANE CURVES ESLAM BADR, FRANCESC BARS, AND ELISA LORENZO To Pilar Bayer for her 70th birthday Abstract. Given a smooth curve defined over a field k that admits a non-singular plane model over k, a fixed separable closure of k, it does not necessarily have a non-singular plane model defined over the field k. We determine under which conditions this happens and we show an example of such phenomenon. Now, even assuming that such a smooth plane model exists, we wonder about the existence of non-singular plane models over k for its twists. We characterize twists possessing such models and use such characterization to improve, for the particular case of smooth plane curves, the algorithm to compute twists of non-hyperelliptic curves wrote recently down by the third author. We also show an example of a twist not admitting such non-singular plane model. As a consequence, we get explicit equations for a non-trivial Brauer-Severi surface. Finally, we obtain a theoretical result to compute all the twists of smooth plane curves with cyclic automorphism group having a k-model whose automorphism group is generated by a diagonal matrix. Some examples are also provided. 1. Introduction Let C be a smooth curve over k, i.e. C a projective non-singular and geometrically irreducible curve defined over a field k. Denote by C the curve C ×k k where k is a fixed separable closure of k, and by Aut(C) its automorphism group. We assume, once and for all, that C is non-hyperelliptic of genus g ≥ 3. With the method exposed in [16] we can compute the twists of C; i.e. a smooth curve C ′ over k with a k-isomorphism ϕ : C → C ′ . The set of twists of C modulo k-isomorphism, denoted by Twistk (C), is in one to one correspondence with the first Galois cohomology set H1 (Gal(k/k), Aut(C)). Given a cocycle ξ ∈ H1 (Gal(k/k), Aut(C)), the idea behind computing equations for the twist, is finding a Gal(k/k)-modulo isomorphism between the subgroup generated by the image of ξ in Aut(C) with a subgroup of a general linear group GLn (k). After that, by making explicit Hilbert’s 90 Theorem, we can compute an isomorphism ϕ : C → C ′ over k such that ξτ = ϕ−1 ·τ ϕ for all τ of the Galois group Gal(k/k), and hence, obtain equations for the twist. For non-hyperelliptic curves, see a description in [15], the canonical model gives a natural inclusion Aut(C) ,→ PGLg (k) and with the natural representation of Aut(C) on the regular differential forms, Aut(C) will live inside a general linear group (then so does the subgroup generated by the image of a 1-cocycle in PGLg (k) provided by the canonical model), Now consider a smooth plane curve C over k, i.e. C is a smooth curve over k that admits a non-singular plane model over k. Therefore, C has a gd2 complete linear series which defines a map Υ : C ,→ P2k , where P2k is the 2-th projective space over k, and moreover Image(Υ) is defined by the zeroes of a degree d polynomial in X, Y, Z with coefficients in k. Denote such a model by FC (X, Y, Z) = 0, in particular g = 12 (d − 1)(d − 2). It is well-known that the complete linear series gd2 is unique up to conjugation in P GL3 (k), the automorphism group of P2k , see [10, Lemma 11.28]. Therefore, any k-model of C is defined by FP C (X, Y, Z) := F (P (X, Y, Z)) = 0 for some P ∈ PGL3 (k). Furthermore, a plane model of C is defined over k if there exists a Q ∈ PGL3 (k) such that FQC (X, Y, Z)∈ k[X, Y, Z]. The group Aut(C) is isomorphic via Υ to the automorphism group Aut(FP C ) of any of its k-models, and all these groups are conjugate in PGL3 (k). We say that C admits a smooth plane model over k if it is k-isomorphic to a smooth plane model defined over k. The aim of this paper is to make a study of the twists for smooth plane curves by considering the embedding Aut(C) ,→ PGL3 (k) instead the one given by the canonical model. The embedding is of Gal(k/k)-groups if C admits a smooth plane model over k. 2010 Mathematics Subject Classification. 11G30, 11D41,14H37, 14H50, 14H45, 12F12. Key words and phrases. non-singular plane curves; automorphism groups; twist. E. Badr and F. Bars are supported by MTM2013-40680-P. 1 2 E. BADR, F. BARS, AND E. LORENZO This approach leads also to two natural questions: the first one, given C a smooth plane curve defined over a field k, does it admit a smooth plane model defined over the base field k; and secondly, if the answer is yes, does every twist of C over k also have a smooth plane model defined over k? For both questions the answer is no in general, it does not. We obtain general results for the curves and the twists for which the above questions always have an affirmative answer, and also we show different examples concerning the negative general answer. Interestingly, in the way to get these examples, we need to handle with non-trivial Brauer-Severi surfaces, and we are able to compute explicit equations of a non-trivial one. Moreover, for smooth plane curves defined over k with a cyclic automorphism group generated by a diagonal matrix, we obtain a general theoretical result to obtain all its twists. Families of such smooth plane curves have already been studied by the first two authors in [2, 3]. These families have genus arbitrarily big, so the method in [16] does not work for them. 1.1. Outline. The structure of this paper is as follows. In Section 2, we introduce the background and basic definition about Galois cohomology and Brauer-Severi varieties needed in the sequel. We devote Section 3 to deal with the question on the field where there exists a non-singular model for a smooth plane curve C defined over k. We prove that if the degree of a non-singular plane model of C is coprime with 3 or C has a k-point or the 3-torsion of the Brauer group of k is trivial (in particular if k is a finite field), then the curve C admits a smooth plane model over k: Theorem 3.6 and Corollaries 3.2, 3.3. Moreover, we prove that a smooth plane model of C always exists in a finite field extension of k of degree dividing 3, see Theorem 3.5. We end Section 3 with an explicit example of a smooth plane curve defined over Q and not admitting a smooth plane model over Q; however, we construct a smooth plane model over a degree 3 extension of Q. In Section 4, we assume that C is a smooth plane curve defined over k having a smooth plane model over k. We obtain Theorem 4.1 characterizing the twists of C having also a smooth plane model over k. Moreover, we construct a family of examples over k = Q where a twist of C over Q does not admit a non-singular plane model over Q (this construction is not explicit because we do not provide equations of such twists). We dedicate Section 5 to detail an explicit example of a smooth plane curve defined over the field Q(ζ3 ) having a twist that does not possess such a model in the field Q(ζ3 ), where ζ3 is a primitive 3rd root of unity. Interestingly, we find the already mentioned explicit equations for a non-trivial Brauer-Severi variety. In Section 6, we study the twists for smooth plane curve C over k with a non-singular plane model over k, FP C ∈ k[X, Y, Z] such that Aut(C) is a cyclic group. We prove that if Aut(FP C ) is represented in PGL3 (k) by a diagonal matrix, then all the twists are diagonal, i.e. of the form FP DC (X, Y, Z) = 0 with D a diagonal matrix, Theorem 6.2. We apply such result to a family of curves (one for each degree) where the techniques of [16] does not apply if the degree is too big, see Theorems 6.4, 6.5. In the case that C does not admit such a model where Aut(FP C ) is cyclic diagonal in PGL3 (k), we also construct an example where not all the twists are given by a diagonal twist. Finally, we apply the algorithm in [16] to the simplest degree 5 example in section 6, this shows the improve- ments of Theorem 6.4 and Theorem 6.2 to compute the twists. 1.2. Notation and conventions. We set the following notations, to be used throughout. By k we denote a field, k is a separable closure of k and L is an extension of k inside k. By ζn we always mean a fixed primitive n-th root of unity inside k when the characteristic of k is coprime with n. We write Gal(L/k) for the Galois group of L/k. All the Galois groups in this paper, when acting on sets, we denote it by left exponentiation. We write H i (Gal(L/k), N ) with i ∈ {0, 1} for the Galois cohomology set of a Gal(L/k)-group N . In the particular case L = k, we also denote Gk instead of Gal(k/k) and H 1 (k, N ) instead of H 1 (Gk , N ). Br(k) denotes the Brauer group of k consisting of the central simple algebras over k modulo k-algebras isomorphism. When we work with groups, we use the SmallGroup Library-GAP [7]. Where the group < N, r > or GAP (N, r) denotes the group of order N that appears in the r-th position in such library. By ID(G), we mean the corresponding GAP notation for the group G. For cyclic groups, we use the standard notation Z/nZ. ON TWISTS OF SMOOTH PLANE CURVES 3 By smooth curve over k we mean a projective, non-singular and geometrically irreducible curve defined over k, and usually we denote it by C or Ck . As usual C corresponds to C ×k k, (the curve C over k), Aut(C) the automorphism group, and g(C) denotes the genus of the curve, and once and for all we assume that g(C) ≥ 2. We denote by Prk the r-th projective space over the field k. By a smooth plane curve C over k we mean a smooth curve over k, which admits a non-singular plane model FC (X, Y, Z) = 0 over k of degree d ≥ 4. We denote by FP C (X, Y, Z) = 0 with P ∈ PGL3 (k) another non- singular model, where FP C (X, Y, Z) := FC (P (X, Y, Z)). By Aut(FP C ) we mean the automorphism group of the curve FP C (X, Y, Z) = 0 in P2k which is a finite subgroup of PGL3 (k). We have that Aut(FP C ) = P −1 Aut(FC )P as subgroups of PGL3 (k). A linear transformation A = (ai,j ) of P2 is also written as [a1,1 X + a1,2 Y + a1,3 Z : a2,1 X + a2,2 Y + a2,3 Z : a3,1 X + a3,2 Y + a3,3 Z]. Given C a smooth plane curve over k, we say that C admits a non-singular plane model over L if there exists P ∈ PGL3 (k) such that FP C (X, Y, Z) ∈ L[X, Y, Z], and C and FP C (X, Y, Z) = 0 are isomorphic over L. By an abuse of language, if a smooth plane curve C over k admits a non-singular plane model over k given by FP C = 0, we identify C with the plane model FP C = 0 and we identify Aut(C) with Aut(FP C ) as a fixed finite subgroup of PGL3 (k). Acknowledgments. The authors would like to thank Xavier Xarles for his suggestions and comments con- cerning our work and to clarify different aspects of his work in [19]. We are very happy also to thank René Pannekoek for e-mailing us his Master thesis [18] and comments on Brauer-Severi surfaces related to his Master thesis. Finally, we thank Christophe Ritzenthaler who pointed us different comments concerning smooth plane curves over finite fields an hyperelliptic curves. 2. Galois cohomology and Brauer-Severi varieties In this section, we state different results about Galois cohomology to be used through the paper (see a general approach in [20, Chapter III]). Definition 2.1. Given C a smooth curve over k. A twist of C over k is a smooth curve C ′ defined over k such that C ∼=k C ′ . Given two twists of C over k, namely C1 , C2 , we say that they are equivalent if there exist an isomorphism C1 ∼ = C2 defined over k. The set of twists of C over k modulo the above equivalence is denoted by Twistk (C). The following result is well-known. Theorem 2.2. There exists a bijection between the sets Twistk (C) and H 1 (k, Aut(C)). Recall that the above bijection [C ′ ] 7→ [ξ] sends a twist φ : C → C ′ to the cocycle ξ : τ 7→ ξτ := φ−1 · τ φ ∈ Aut(C) where τ ∈ Gk . Lemma 2.3. Let C be a curve over k admitting a plane model FC = 0 over k̄. Let us assume that there exists a matrix P ∈ P GL3 (k) with FP C (X, Y, Z) ∈ k[X, Y, Z]. Then, there exists a twist C ′ of C over k given by the non-singular plane model FP C = 0 over k. Furthermore, we have a map Σ′ : Twistk (C) → H 1 (k, PGL3 (k)). Proof. Let h : C → FP C be a k- isomorphism, and consider the 1-cocycle σ 7→ h−1 ·σ (h) ∈ Aut(C) in H 1 (k, Aut(C)) = T wistk (C). Then there exists a twist λ : C → C ′ of C over k such that h−1 · σ h = λ−1 · σ λ for all σ ∈ Gk . Therefore λ ◦ h−1 : FP C → C ′ is an isomorphism defined over k. Now we can identify Aut(C ′ ) with Aut(FP C ) as Gk -groups. Since Aut(FP C ) has an injective representation inside Aut(P2k ) = PGL3 (k) as a Gk -group, we get a natural map in Galois cohomology Σ′ : T wistk (C) = H 1 (k, Aut(C ′ )) → H 1 (k, P GL3 (k)).  Definition 2.4. A Brauer-Severi variety D over k of dimension r is a smooth projective variety such that the variety D ⊗k k over k is isomorphic to the projective space Prk of dimension r over k. 4 E. BADR, F. BARS, AND E. LORENZO The following result is also well-known [12, Corollary 4.7]: Lemma 2.5. The Brauer-Severi varieties over k of dimension r, up to k-isomorphism, are in bijection with H 1 (Gal(k/k), P GLr+1 (k)) = H 1 (Gal(k/k), Autk (Prk )). Denote by Azkn the set of all isomorphic classes of central simple algebras A of dimension n2 over k (they split in a separable extension of degree n of k). The next result is Corollary 3.8 in [12]. Theorem 2.6. Let k be a field, then there exists a bijection between the following sets Azkn ←→ H1 (Gk , PGLn (k)). For completeness, we recall the map defining the above bijection: given A ∈ Azkn , we always get a Galois extension L/k of degree n and an isomorphism f : A ×k L → Mn (L) such that there exists a matrix Aτ ∈ Aut(Mn (L)) = P GLn (L) for every τ ∈ Gal(L/k) that makes the following diagram commutative: A ×k L →f Mn (L) τ↑ ↑ Aτ ◦ τ A ×k L →f Mn (L) The map τ 7→ Aτ defines an element of H 1 (Gal(L/k), PGLn (L)), and we obtain an element in H 1 (k, PGLn (k)) through the inflation map. Definition 2.7. Let L/k be a cyclic extension of degree n with Gal(L/k) = ⟨σ⟩, and fix an isomorphism χ : Gal(L/k) → Z/nZ. Given a ∈ k ∗ , we consider (a, χ), the n-dimensional vector space over L with basis 1, e, . . . , en−1 , i.e. (χ, a) := ⊕1≤i≤n−1 Lei , where the multiplication rules are given by eλ = σ(λ)e for λ ∈ L and en = a. Such (χ, a) becomes a central simple algebra of dimension n2 over k which splits in L (see [23, §2]), and is called a cyclic algebra of k. Theorem 2.8. All the elements of Azk3 are cyclic algebras of the form (χ, a) as in Definition 2.7 with n = 3. In particular, modulo isomorphism of k-algebras, (χ, a) ∈ Azk3 is the trivial k-algebra if and only if a is a norm of L/k of an element of L. Moreover, the assignment (χ, a) ∈ Azk3 7→ inf ({Aτ }τ ∈Gal(L/k) ) ∈ H 1 (k, PGL3 (k))   00a   is given by an Aσ of the shape  1 0 0  . Here inf denotes the inflation map in Galois cohomology. 01 0 Proof. By definition, any central cyclic simple algebra of Azk3 can be expressed as described.  Recallalso that 0 0a   the map f : A ⊗k L → M3 (L) is given by f (λ ⊗ 1) = diag(λ, σ(λ), σ (λ)) and f (e ⊗ 1) =  1 0 0 . It is a 2 0 1 0 k result of Wedderbum [25], that all elements of Az3 corresponds to cyclic algebras. Moreover, it is well-known that a cyclic algebra (χ, a) is trivial if and only if a is a norm of its splitting field. The last statement concerning the 1-cocycle assignment follows by [23, Example 5.5], after defining first the cocycle element in H 1 (Gal(L/k), PGL3 (L)) and using [12, Lemma 3.7] for the inflation map in Galois cohomology groups.  Because the Brauer group of a finite field is trivial, and taking cohomology of the short exact sequence ∗ 1 → k → GLn (k) → P GLn (k) → 1, we mention: Lemma 2.9. Let k be a finite field, then H1 (Gk , PGLn (k)) = 1. ON TWISTS OF SMOOTH PLANE CURVES 5 3. The field of definition of a non-singular plane model In this section, we prove that if a smooth curve defined over k admits a non-singular plane model over k, then it is always possible to find a non-singular plane model defined over an extension L/k of degree dividing 3. Moreover, for a particular field k or if the smooth plane model has degree coprime with 3, we prove that we can always find a non-singular plane model defined over the base field k. We provide an example of a curve defined over Q that does not admit a smooth plane model over Q, but that it does over a Galois extension of Q of degree 3. Roé and Xarles prove the following result in [19, Corollary 6]. Theorem 3.1 (Roé-Xarles). Let C be a smooth projective curve defined over k such that C has a non-singular plane model. Let Υ : C ,→ P2k be a morphism given by the uniqueness of the gd2 -linear system over k, then there exists a Brauer-Severi variety D (of dimension two) defined over k, together with a k-morphism g : C ,→ D such that g ⊗k k : C → P2k is equal to Υ. From the above result, one obtain remarkable consequences. Corollary 3.2. Let C be a smooth curve over k that admits a non-singular plane model. Assume that C has a k-point, i.e. C(k) is not-empty. Then C admits a non-singular plane model over k. Proof. By a Severi result, see [12, Prop.4.8], a Brauer-Severi variety over k of dimension n with a k-point is isomorphic over k to Pnk . By Theorem 3.1, the map g : Ck → D ∼ = P2k defined over k defines the non-singular plane model of C over k.  Corollary 3.3. Consider a field k such that Br(k)[3] is trivial, where Br(k)[3] denotes the 3-torsion of Br(k). Then any smooth plane curve C over k, admits a non-singular plane model over k, and in particular any twist of C over k admits also a non-singular plane model over k. Proof. A non-trivial Brauer-Severi surface over k corresponds to a non-trivial 3-torsion element of Br(k), there- fore is such group is empty, by Theorem 3.1 the g2d -system factors through g : Ck ,→ P2k and all of them are defined over k, so, they define a plane model of C over k.  Remark 3.4. For a finite field k (see Lemma 2.9) or k = R, it is known that Br(k)[3] is trivial, therefore any smooth plane curve over such fields admits always a non-singular plane model over k. Theorem 3.5. Let C be a smooth plane curve defined over k, then it admits a non-singular plane model over L such that [L : k] | 3, i.e. ∃P ∈ PGL3 (k) such that FP C ∈ L[X, Y, Z] and such that C and FP C = 0 are L-isomorphic. Proof. From Theorem 3.1, we have a k-morphism of C to a Brauer-Severi surface D over k. By Theorem 2.6, D corresponds to a central simple algebra over k of dimension 9 which splits (if is not the trivial algebra) in at degree 3 Galois extension L of k, therefore D ⊗k L corresponds to the trivial element in H 1 (Gal(k/L), PGL3 (k)), by theorem 2.6. Thus, D ⊗k L ∼ = P2L over L. We then obtain that g ⊗k L : C ⊗k L ,→ P2L are all defined over L, thus we have a non-singular plane model of C over L. Lastly, because all the non-singular plane models of C over k are of the form FP C (X, Y, Z) = 0 for some P ∈ PGL3 (k), we deduce the result.  The next result is a particular case of an argument by Roé and Xarles in [19] following Châtelet [6]. Theorem 3.6. Let C be a smooth curve defined over k, such that admits a non-singular plane model of degree d with d coprime with 3. Then C admits a non-singular plane model over k. Proof. By the results of the previous section, Brauer-Severi surfaces over k corresponds to elements of H 1 (k, PGL3 (k)), hence to the set of equivalence classes of central simple algebras of dimension 9 with a splitting field of degree 3 over k (thus, they are elements of the Brauer group Br(k) of the field k of order dividing 3). 6 E. BADR, F. BARS, AND E. LORENZO Moreover, if D is a Brauer-Severi surface over a field k, then its class [D] in the Brauer group Br(k) verifies that in the exact sequence P ic(D) → P ic(D ⊗k k) ∼ = Z → Br(k), the last map send 1 to [D], and hence the image of some generator of P ic(D) is equal to m, where m is the order of [D]. Consequently, m divides 3, as the order of [D] does. Now, if C is a curve over k in P ic(D) such that C has a non-singular plane model of degree d, then the image of C in P ic(B ⊗k k) ∼ = Z is equal to the degree d. Therefore, if d is coprime with 3, we thus get m = 1, and D is the projective plane P2k (see [19, Theorem 13] for a more general statement on hypersurfaces in Brauer-Severi varieties).  Corollary 3.7. Let C be a smooth curve defined over k which admits a non-singular plane model over k of degree d, coprime with 3. Then, every twist C ′ ∈ Twistk (C) admits a non-singular plane model over k. Proof. It follows, by our assumption, that every twist of C over k admits a non-singular plane model over k of degree d, coprime with 3. Hence, non-singular plane models over k exist for twists of C over k, by Theorem 3.6.  3.1. An example of a smooth plane curve over Q without a non-singular plane model over Q. Let us consider Qf the splitting field of the polynomial f (x) = x3 +12x2 −64. It is an irreducible polynomial and the discriminant of f is (26 32 )2 , then Gal(Qf /Q) ≃ Z/3Z, moreover by a computation in SAGE, the discriminant of the field Qf is a power of 3, and the prime 2 becomes inert in Qf . Let us denote the roots of f by a, b, c in a fixed algebraic closure of Q, and let us call σ the element in the Galois group that acts by sending a → b → c. Proposition 3.8. The smooth plane curve over Qf ab 3 3 C : 64x6 + aby 6 + az 6 + 8x3 y 3 + y z + az 3 x3 = 0, 8 has Q as a field of definition, but it does not admit a plane non-singular model over Q. Proof. The matrix   00 2   ϕ = 1 0 0 01 0 defines an isomorphism ϕ : σ C → C. This isomorphism ϕ satisfies the Weil cocycle condition [26] (ϕσ3 = ϕ3σ = 1), we therefore obtain that the curve is defined over Q, and that there exists an isomorphism φ0 : CQ → C where CQ is a rational model such that ϕ = φ0 σ φ−1 0 ∈ PGL3 (Q). The assignation ϕτ := φ0 φ0 τ −1 defines an element of H 1 (Gal(Qf /Q), PGL3 (Qf )), by Theorem 2.8, this cohomology element is non-trivial because 2 is not a norm of an element of Qf (since 2 is inert in Qf ). Therefore φ0 is not given by an element of PGL3 (Qf ), or of PGL3 (Q) because the cohomology class by the inflation map is neither trivial. Therefore the curve C over Q does not admits a non-singular plane model over Q (because if admitted a non-singular plane model over Q, such model would be of the form FP QC (X, Y, Z) = 0 for some P ∈ PGL3 (k) where FQC (X, Y, Z) = 0 a non-singular model over Qf , therefore φ0 would be representative by P ∈ PGL3 (k) which is not).  Remark 3.9. We have just seen an example of a curve defined over a field k not admitting a particular model (a plane one) over the same field. For hyperelliptic models, we find such examples after Proposition 4.14 in [13]. In [11, chp. 5,7], there are also examples of hyperelliptic curves and smooth plane curves where the field of moduli is not a field of definition, so, in particular, there are not such models defined over the fields of moduli. 4. On twists of plane models defined over k In this section, we assume, once and for all, that C is a smooth curve defined over k with a non-singular plane model also defined over k, i.e. we can assume that C is given by an equation FC = 0 with FC ∈ k[X, Y, Z]. We provide results characterizing when all the twists of C admit a non-singular plane model over k, and we give a (non-explicit) example of a family of such curves C having twists not admitting a plane model over k. ON TWISTS OF SMOOTH PLANE CURVES 7 Theorem 4.1. Let C be a curve defined over a field k with a plane non-singular model FC (X, Y, Z) = 0 defined over k. Then there exists a natural map Σ : H 1 (k, Aut(FC )) → H 1 (k, PGL3 (k)), defined by the inclusion Aut(FC ) ⊆ PGL3 (k) as Gk -groups. The kernel of Σ is the set of all twists of C that admit a non-singular plane model over k. Moreover, any such a plane model is obtained through an automorphism of P2 , that is, it is of the form FM C (X, Y, Z) := FC (M (X, Y, Z)) ∈ k[X, Y, Z] for some M ∈ PGL3 (k). Proof. The map is clearly well-defined. If a twist C ′ admits a non-singular plane model FC ′ over k, the isomorphism from FC ′ to FC is then given by an element M ∈ PGL3 (k) (as any isomorphism between two non- singular plane curves of degrees > 3 is given by a linear transformation in P2 , [5]). Hence, the corresponding 1-cocycle σ 7→ M σ M −1 ∈ Aut(FC ) becomes trivial in H 1 (k, PGL3 (k)). Conversely, if a twist C ′ is mapped by Σ to the trivial element in H 1 (k, PGL3 (k)), then this twist is given by a k-isomorphism φ : FC → C ′ defined by a matrix M ∈ PGL3 (k) that trivializes the cocycle and such an M produces a non-singular plane model defined over k.  Remark 4.2. We can reinterpret the map Σ in Theorem 4.1 as the map that sends a twist C ′ to the Brauer- Severi variety D in Theorem 3.1. Remark 4.3. Consider a smooth plane curve C defined over k, and assume that it has degree d coprime with 3 or that Br(k)[3] is trivial, then Σ in Theorem 4.1 is the trivial map by Corollaries 3.7 and 3.3. Remark 4.4. Theorem 4.1 can be used to improve the algorithm for computing twists for non-hyperelliptic curves, see [16] or [15, Chp.1], for the special case of non-singular plane curves. The algorithm requires to compute a canonical model in Pg−1 , solutions to Galois embedding problems, and constructing equations for the twists. The last step needs to see Aut(C) ⊆ GLg (k̄) via the canonical embedding and the action of the automorphism group on the vector space of regular differentials, so, we can use an explicit version of Hilbert 90 Theorem. Now, if Σ is trivial in Theorem 4.1, then we can work on P2 instead of on Pg . The exact sequence 1 → k̄ → GL3 (k̄) → PGL3 (k̄) gives the exact, well-defined sequence 1 → H1 (k̄, GL3 (k̄)) → H1 (k̄, PGL3 (k̄)). Hence, the trivial element of H1 (k̄, PGL3 (k̄)) corresponds to an element in H1 (k̄, GL3 (k̄)) and we can proceed again by Hilbert 90 Theorem. In the appendix, we use this improvement to compute the twists of some particular families of plane curves over k having a cyclic diagonal automorphism group. To finish we construct a family of smooth curves defined over Q that admits a non-singular plane model over Q but some of its twists do not admit a non-singular plane model over Q. This construction is not explicit in the sense that we do not construct the equations of the twist and the Brauer-Severi surface, see next section for an explicit construction giving defining equations. Theorem 4.5. Let p ≡ 3, 5 mod 7 be a prime number. Take a ∈ Q with a ̸= −10, ±2, −1, 0. Consider the family Cp,a of smooth plane curves over Q given by 1 6 1 a Cp,a : X 6 + Y + 4 Z 6 + 3 (p2 X 3 Y 3 + pX 3 Z 3 + Y 3 Z 3 ) = 0. p2 p p Then, there exists a twist C ′ ∈ TwistQ (Cp,a ) which does not admit a non-singular plane model over Q. We need some lemmas before proving Theorem 4.5. Lemma 4.6. Consider the family of non-singular plane curves over Q defined by the equation Ca : X 6 + Y 6 + Z 6 + a(X 3 Y 3 + X 3 Z 3 + Y 3 Z 3 ) = 0, with a ̸= −10, ±2, −1, 0. The full automorphism group Aut(Ca ) is generated by S := [X; ζ3 Y ; ζ32 Z], U := [X; Y ; ζ3 Z], T := [Z; X; Y ], and R := [Y ; X; Z]. In particular, it is isomorphic to GAP (54, 5). 8 E. BADR, F. BARS, AND E. LORENZO Proof. First, it is necessary that a ̸= ±2, −1 for non-singularity of Ca . Second, we use similar techniques and notation to the ones used and developed in [2], in particular Aut(Ca ) should be one of the groups in [2, Theorem 2]. We have ⟨R, S, T, U ⟩ ≤ Aut(Ca ), hence Aut(Ca ) is not conjugate to a cyclic group, the Klein group P SL(2, 7), the icosahedral group A5 , the alternating group A6 (i.e. Ca is not K-equivalent to the Wiman sextic curve), or the Hessian groups Hess∗ with ∗ ∈ {36, 72}. On the other hand, Ca is not a descendant of the degree 6 Klein curve K6 , since 54 - |Aut(K6 )|(= 63), also Aut(Ca ) fixes no points in the projective plane P2 , thus from Mitchel’s classification that is explicitly explained in [2], Aut(Ca ) fixes a triangle. Consequently, we need only to worry about the following situations: Aut(Ca ) is conjugate to the Hessian group Hess216 or to a subgroup of Aut(F6 ) (i.e. Ca is a descendant of the Fermat curve of degree 6). We note that there is a unique representation of the Hessian group Hess216 inside PGL3 (K), up to conjuga- tion. Thus, without loss of generality, we consider Hess216 = ⟨S, T, U, V ⟩ where   1 1 1   V =  1 ζ3 ζ32  . 1 ζ32 ζ3 Now, if Aut(Ca ) = P −1 · ⟨S, T, U, V ⟩ · P for some P ∈ PGL3 (k), then it is easy to check that it must be Aut(Ca ) = ⟨S, T, U, V ⟩. Moreover, if V ∈ Aut(Ca ) then a = −10 (see the coefficient of X 4 Y Z), which is not our case. Let us now inspect the automorphism group of the Fermat curve of degree 6. Aut(F6 ) = ⟨[ζ6 X; Y ; Z], [X; ζ6 Y ; Z], [Y ; Z; X], [X; Z; Y ] ⟩ ≃ GAP (216, 92), where ζ6 is a primitive 6-th root of unity. This group contains a unique group of order 54, up to conjugation. Hence, we may assume that Ca is a descendant of the Fermat curve F6 : X 6 + Y 6 + Z 6 = 0 through a projective transformation P ∈ PGL3 (K) such that P −1 ⟨S, U, T, R⟩P = ⟨S, U, T, R⟩. Consequently, the transformed equation should be again of the form CP : X 6 + Y 6 + Z 6 + a′ (X 3 Y 3 + X 3 Z 3 + Y 3 Z 3 ) for some a′ ∈ Q. Finally, elements of Aut(F6 ) are of the forms ′ ′ ′ ′ ′ [X; ζ6r Y ; ζ6r Z], [X; ζ6r Z; ζ6r Y ], [ζ6r Y ; X; ζ6r Z], [ζ6r Z; ζ6r Y ; X], [ζ6r Y ; ζ6r Z; X], [ζ6r Z; X; ζ6r Y ] with r, r′ ∈ Z. Because a ̸= 0, then any of these forms belongs to Aut(CP ) only if 2|r, r′ . This gives exactly 6(3 × 3) = 54 automorphisms inside Aut(F6 ). As a conclusion, the full automorphism group of Ca with a ̸= −10, ±2, −1, 0 is of order 54 and is isomorphic to GAP (54, 5).  Remark 4.7. If the automorphism group of a smooth plane curve, as a subgroup of PGL3 (k) contains a subgroup conjugate to ⟨S, U, T, R⟩, then the degree d of the plane model is divisible by 3. This follows because the non-singularity plus having the automorphism T implies that we must have one of the following cores 1: (1) X d + Y d + Z d , (2) X d−1 Y + XZ d−1 + Y d−1 Z, (3) X d−1 Z + Y Z d−1 + XY d−1 But also we have U as an automorphism of the plane model, therefore the plane model could only have the core X d + Y d + Z d with 3|d. Lemma 4.8. Given a ̸= −10, ±2, −1, 0 and α0 ∈ Q, the family of curves 1 6 1 a Cα0 ,a : X 6 + 2 Y + 4 Z 6 + 3 (α02 X 3 Y 3 + α0 X 3 Z 3 + Y 3 Z 3 ) = 0, α0 α0 α0 has automorphism group isomorphic to GAP (54, 5) with [Y, Z, α0 X] ∈ Aut(Cα0 ,a ), and the elements of Aut(Cα0 ,a ) √ are defined over the field Q(ζ3 , 3 α0 ), where ζ3 is a primitive 3-rd root of unity. 1For a non-zero monomial cX i Y j Z k , its exponent is defined to be max{i, j, k}. For a homogeneous polynomial F , the core of F is the sum of all terms of F with the greatest exponent. ON TWISTS OF SMOOTH PLANE CURVES 9 Proof. The curve Cα0 ,a and Ca of Lemma 4.6 are k-isomorphic through the projective transformation P = diag(1; β; µ) where β 3 = α10 , and µ3 = α12 .  0 Proof. (of Theorem 4.5) √ Consider the Galois extension M/Q with M = Q(cos(2π/7), ζ3 , 3 p) where all the elements of Aut(Cp,a ) are defined. Let σ be a generator of the cyclic Galois group Gal(Q(cos(2π/7))/Q). We define a 1-cocycle in √ Gal(M/Q) ∼ = Gal(Q(cos(2π/7))/Q) × Gal(Q(ζ3 , 3 p)/Q) to Aut(Cp,a ) by mapping (σ, id) 7→ [Y, Z, pX] and (id, τ ) 7→ id. This defines an element of H 1 (M/Q, Aut(Cp,a )). Consider its image by Σ inside H 1 (M/Q, PGL3 (M )). We need to check that its image is not the trivial element, and then the result is an immediate consequence by Theorem 4.1. By Theorem 2.6, H 1 (M/Q, PGL3 (M )) is the set of central simple algebras over Q of dimension 9 which splits in a degree 3 field inside M . If we consider the image in H 1 (Gal(Q(cos(2π/7))/Q), PGL3 (Q(cos(2π/7)))) then it is non-trivial if and only if p is not a norm of the field extension Q(cos(2π/7))/Q. By [24, Theorem 2.13], the ideal (p) is prime in Q(cos(2π/7))/Q, therefore p is not a norm of an element of Q(cos(2π/7)). Now H 1 (M/Q, PGL3 (M )) is the union of the above central simple algebras over Q running through the subfields F ⊂ M of degree 3 over Q, see [12]. Thus the element is not trivial, which was to be shown.  5. Explicit non-plane model twists over k of a plane model defined over k In this section, we explicitly construct a twist of a curve Ca over Q(ζ3 ) which does not admit a plane model over Q(ζ3 ). In particular, we construct a non-trivial Brauer-Severi surface over Q(ζ3 ) giving its equations inside P9Q(ζ3 ) . Let us consider the curve Ca : X 6 + Y 6 + Z 6 + a(X 3 Y 3 + Y 3 Z 3 + Z 3 X 3 ) = 0 defined over a number field k ⊇ Q(ζ3 ) where ζ3 is a primitive third root of unity and a ∈ k. For a ̸= −10, −2, −1, 0, 2, it is a non- hyperelliptic, non-singular plane curve of genus g = 10 and its automorphism group is the group of order 54 determined in the previous section. The algorithm in [16], allows us to compute all the twists of Ca , previous computation of its canonical model in P9 . We follow such algorithm, since this time we will see that Σ is not trivial, so we cannot use the improvements in Remark 4.4. 5.1. A canonical model of Ca in P9 . Let us denote by αi the six different root of the polynomial T 6 +aT 3 +1 = 0, and define the points on Ca : Pi = (0 : αi : 1), Qi = (αi : 0 : 1) and ∞i = (αi : 1 : 0). The divisor of the function x = X/Z is div(x) = Pi − ∞. Let P = (X0 : Y0 : 1) ∈ Ca , the function x is a uniformizer at P if the polynomial T 6 + a(X03 + 1)T 3 + X06 + aX03 + 1 = 0 does not have double roots. That is, if X06 + aX03 + 1 ̸= 0 or 4(X06 + aX03 + 1) ̸= a2 (X03 + 1)2 . Let us denote by βi the six different roots of the polynomial T 6 + a+2 2a T 3 +1 = 0 √ and denote by Vij = (βi : ζ3j+1 3 − a2 (βi3 + 1) : 1) where j ∈ {1, 2, 3}. In order to compute ordP (dx) we need to use the expression y 2 2y 3 + a(x3 + 1) dx = − 2 3 dy x 2x + a(y 3 + 1) for the points Qi and Vi,j . Notice that div(2y 3 + a(x3 + 1)) = Vi,j − 3∞. For the points at infinity, we use that the degree of a differential is 2g − 2 = 18. We finally get div(dx) = 2Qi + Vi,j − 2∞′ s. Hence, a basis of regular differentials is given by xdx x2 y2 1 ω1 = ω = , ω2 = ω, ω3 = ω, ω4 = ω y(2y 3 3 + a(x + 1)) y x xy y 1 x 1 ω5 = xω, ω6 = ω, ω7 = ω, ω8 = yω, ω9 = ω, ω10 = ω. x y y x We list the divisors of these differentials below. div(ω1 ) = Pi + Qi + ∞, div(ω2 ) = 3Pi , div(ω3 ) = 3Qi , div(ω4 ) = 3∞ 10 E. BADR, F. BARS, AND E. LORENZO div(ω5 ) = 2Pi + Qi , div(ω6 ) = 2Qi + ∞, div(ω7 ) = Pi + 2∞, div(ω8 ) = Pi + 2Qi , div(ω9 ) = 2Pi + ∞, div(ω10 ) = Qi + 2∞. Lemma 5.1. The ideal of the canonical model of Ca in P9 [ω1 , ω2 , ω3 , ω4 , ω5 , ω6 , ω7 , ω8 , ω9 , ω10 ] is generated by the polynomials ω4 ω9 = ω72 , ω4 ω6 = ω10 2 , ω4 ω1 = ω7 ω10 , ω4 ω5 = ω9 ω10 , ω4 ω8 = ω6 ω7 , ω4 ω2 = ω7 ω9 , ω4 ω3 = ω6 ω10 , ω3 ω10 = ω62 , ω2 ω7 = ω92 , ω6 ω9 = ω12 , ω3 ω5 = ω82 , ω2 ω3 = ω5 ω8 , ω2 ω8 = ω52 , ω22 + ω32 + ω42 + a(ω5 ω8 + ω6 ω10 + ω7 ω9 ) = 0. We denote by Ca this canonical model. Proof. If ω4 ̸= 0, then the des-homogenization of this ideal with respect to ω4 gives the affine curve Ca for Z = 1. If ω4 = 0, then ω7 = ω10 = 0, so ω6 = ω9 = 0 and ω1 = 0, so if ω3 ̸= 0 we recover the part at infinity (Z = 0) of Ca . If ω4 = ω3 = 0, then all the variables are equal to zero which produces a contradiction. To check that it is non-singular, we need to see if the rank of the matrix of partial derivatives of the previous generating functions has rank equal to 8 = dim(P9 ) − dim(C) at every point. If ω4 ̸= 0, then the partial derivatives of the first seven equation plus the last one produce linearly independent vectors If ω4 = 0, we have already seen that ω3 ̸= 0 and by equivalent arguments, neither it is ω2 . Then the 6th, 7th, 8th, 9th equations plus the last four equations produce the linearly independent vectors.  Remark 5.2. The canonical embedding of Ca in Pg−1 = P9 coincides with the composition of the g2d -linear system of Ca with the Veronese embedding given by: P2 ,→ P9 : (x : y : z) → (xyz : x3 : y 3 : z 3 : x2 y : y 2 z : z 2 x : xy 2 : x2 z : yz 2 ). In particular, we get that the ideal defining the projective space P2 in P9 by the Veronese embedding is generated by the polynomials defined in Lemma 5.1 after removing the last one. 5.2. The automorphism group of Ca in P9 . Let us consider the automorphisms of the curve Ca given by R = [y; x; z], T = [z; x; y] and U = [x; y; ζ3 z]. We easily check that < R, T, U >⊆ Aut(Ca ) and by Lemma 4.6, we obtain that Aut(Ca ) =< R, T, U >. Notice that the pullbacks R∗ (ω) = −ω, T ∗ (ω) = ω and U ∗ (ω) = ζ32 ω. So, in the canonical model, these automorphisms look like     1 0 0 0 0 0 0 0 0 0 1 00 0 0 0 0 0 0 0      0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0     0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0    0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0        0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 R → −R = −  ,T → T =   0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0     0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0    0  0 0 0 0 0  0 0 0 1 0 0 0 0 0  0 0 0 0 1      0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 00 0 0 0 0 1 0 0 and U → ζ32 Diag(1, ζ32 , ζ32 , ζ32 , ζ32 , 1, ζ3 , ζ32 , 1, ζ3 ) = ζ32 U. We define the faithful linear representation Aut(Ca ) ,→ GL10 (k) by sending R, T, U → R, T , U. Moreover, it preserves the action of the Galois group Gk . 5.3. A explicit twist over k = Q(ζ3 ) of Ca without a non-singular plane model over k. Let us consider the subgroup N of Aut(Ca ) generated by N :=< T U >≃ Z/3Z. √ Let us consider the curve Ca defined over k = Q(ζ3 ), and the field extension L = k( 3 7) with Galois √ √ group Gal(L/k) =< σ >≃ Z/3Z, where σ( 3 7) = ζ3 3 7. We define the cocycle ξ ∈ Z1 (Gk , Aut(Ca )) ,→ Z1 (Gk , PGL10 (k)) given by ξσ = T U. Lemma 5.3. The image of the cocycle ξ by the map Σ : H1 (Gk , Aut(Ca )) → H1 (Gk , PGL3 (k)) is not trivial. ON TWISTS OF SMOOTH PLANE CURVES 11 Proof. By construction, the image of the cocycle ξ in H 1 (k, PGL3 (k)) coincides with the inflation of the cocycle in H 1 (Gal(L/k), PGL3 (L)) where ξσ = T U . Now by Theorem 2.8 we conclude, since ζ3 is not a norm in L/k (no new primitive root of unity appears in L than k and ζ3 is not a norm of an element of L).  In order to compute equations defining the twist Ca′ associated to the cocycle ξ (and the Brauer-Severi surface that contains such twist), we need to find a matrix ϕ ∈ PGL10 (k) such that ξσ = ϕ ·σ ϕ−1 . We can then take   1 0 0 0 0 0 0 0 0 0  3 √ √   0 √7 √72 7 0 0 0  3 0 0 0  3   0 7 ζ3 3 72 7ζ32 0 0 0 0 0 0   √ √   0 3 7 ζ 2 3 72 7ζ3 0 0 0 0 0 0   3 √ √   3  0 0 0 0 1 3 7 ζ3 7 2 0 0 √ √ 0  ϕ=  0 0 0 0 1 ζ3 73 3 72 0 0 0   √ √  0 0 0 0 ζ3 3 7 3 72 0 0 0   √ √  0 0 1 ζ3 3 7 ζ3 72  3  0 0 0 0 0 √ √    0 0 0 0 0 0 0 ζ3 ζ3 3 7 3 72  √ √3 0 0 0 0 0 0 0 ζ3 3 7 ζ3 72 If we simply substitute this isomorphism ϕ in the equations of Ca , we will get equations for Ca′ . However, √ even defining a curve over k, this equations are defined over L = k( 3 7). In order to get generators of the ideal defined over k, we use next lemma. √ √ √ √ Lemma 5.4. Let f0 , f1 , f2 ∈ k[x1 , ..., xn ], and define g0 = f0 + 3 7f1 + 72 f2 , g1 = f0 + ζ3 3 7f1 + ζ32 72 f2 , 3 3 √ √ 3 2 3 g2 = f0 + ζ3 7f1 + ζ3 72 f2 . Then the ideals in L[x1 , ..., xn ] generated by < g0 , g1 , g2 > and < f0 , f1 , f2 > are equal. Proof. Clearly, we have the inclusion < g0 , g1 , g2 >⊆< f0 , f1 , f2 >. The reverse inclusion can be checked by √ √ √ writing 3f0 = g0 + g1 + g2 , (ζ3 − 1) 3 7f1 = g1 − ζ3 g2 + (ζ3 − 1)f0 and 72 f2 = g0 − f0 − 3 7f1 .  3 Proposition 5.5. The equations in P9 of the non-trivial Brauer-Severi surface B over k constructed as in Theorem 4.1 from the cocycle ξ above are ω1 ω2 = ζ3 ω5 ω9 + ζ3 ω6 ω8 + 7ζ3 ω7 ω10 , ω22 − 7ω3 ω4 = ζ3 ω5 ω10 + ζ3 ω7 ω8 + ζ3 ω6 ω9 , ω1 ω3 = ω5 ω10 + ζ 2 ω7 ω8 + ζ3 ω6 ω9 , 7ω32 − 7ζ3 ω2 ω4 = ω5 ω9 + ζ32 ω6 ω8 + 7ζ3 ω7 ω10 , 7ω1 ω4 = ζ3 ω5 ω8 + 7ω6 ω10 + 7ζ3 ω7 ω9 , 49ω42 − 7ζ32 ω2 ω3 = ω5 ω8 + 7ζ3 ω6 ω10 + 7ζ32 ω7 ω9 , 2 ω52 + 14ζ3 ω6 ω7 = 7ζ3 ω2 ω10 + 7ω4 ω8 + 7ζ3 ω3 ω9 , ω52 − 7ζ3 ω6 ω7 = 7ω2 ω10 + 7ω4 ω8 + 7ζ32 ω3 ω9 , ω62 + 2ζ3 ω5 ω7 = ζ3 ω2 ω9 + ω3 ω8 + 7ζ3 ω4 ω10 , ω62 − ζ3 ω5 ω7 = ω2 ω9 + ζ3 ω3 ω8 + 7ζ3 ω4 ω10 , 7ω7 + 2ζ3 ω5 ω6 = ζ3 ω2 ω8 + 7ζ3 ω3 ω10 + 7ζ3 ω4 ω9 , 7ω72 − ζ3 ω5 ω6 = ω2 ω8 + 7ω3 ω10 + 7ζ32 ω4 ω9 , 2 2 2 ω82 + 14ζ3 ω9 ω10 = 7ζ32 ω2 ω7 + ω4 ω5 + 7ζ32 ω3 ω6 , ω82 − 7ζ32 ω9 ω10 = 7ζ32 ω2 ω7 + 7ζ32 ω4 ω5 + 7ω3 ω6 , ω92 + 14ζ32 ω8 ω10 = ζ3 ω2 ω6 + ζ32 ω3 ω5 + 7ζ3 ω4 ω7 , ω92 − 7ζ32 ω8 ω10 = ζ32 ω2 ω6 + ζ3 ω3 ω5 + 7ζ3 ω4 ω7 , 2 7ω10 + 2ζ32 ω8 ω9 = ζ32 ω2 ω5 + 7ω3 ω7 + 7ω4 ω6 , 2 7ω10 − ζ32 ω8 ω − 9 = ζ32 ω2 ω5 + 7ζ32 ω3 ω7 + 7ω4 ω6 , Proof. We only need to plug the equations of the isomorphism ϕ into the equations defining Ca and apply Lemma 5.4.  In order to get the equations of the twisted curve, we only need to add the equation that we get by plugging ϕ in ω22 + ω32 + ω42 + a(ω5 ω8 + ω6 ω10 + ω7 ω9 ) = 0, and apply Lemma 5.4 again. Proposition 5.6. The curve Ca′ is a twist over k of the curve Ca for a ̸= −10, −2, −1, 0, 2 which does not admits a non-singular plane model over k and the defining equations of Ca′ in P9 are the ones given in Proposition 5.5 plus the extra equation: ω22 + 14ω3 ω4 + a(ω22 − 7ω3 ω4 ) = 0 12 E. BADR, F. BARS, AND E. LORENZO 5.4. A non-singular plane model for the twist of Ca′ over finite fields. We consider the reductions C̃a and C˜a at a prime p of good reduction of the curve Ca /k and the twist Ca′ /k computed in subsection 5.3. Since k = Q(ζ3 ), the resulting reductions curves are defined over a finite field Fq with q ≡ 1 mod 3, and q = pf for some f ∈ N and p | p. We also assume that p > 21 = (6−1)(6−2)+1 in order to ensure that Aut(C̃a ) ≃< 54, 5 >, see [1, §6] and Lemma 4.6. The natural map Gk → GFq induces a map H1 (k, Aut(Ca )) → H1 (Fq , Aut(C̃a )). Since Z1 (Fq , Aut(C̃a )) ,→ Z (GFq , PGL3 (Fq )) and H1 (GFq , PGL3 (Fq )) = 1, see lemma 2.9, the reduction twist has a non-singular plane 1 models. Clearly, if 7 ∈ F3q , then the twist Ca′ becomes trivial. Otherwise, we get that the reduction of the cocycle ξ is given by its image at π, the Frobenius endomorphism [17], and ξπ can take the values  e 0 0 ζ3e   1 0 0  , 01 0 √ where e = 0, 1, 2 according to the splitting behaviour of the prime p in L = k( 3 7). In the first case, we get the trivial twist. In the later and the former, let assume e = 1 (the other can be treat symmetrically) and q ̸≡ 1 mod 9, we can then take a generator η of Fq3 /Fq , such that η 3 = ζ3 . Then, the cocycle is given (ξσ = ϕ σ ϕ−1 ) by the isomorphism   1 η η2   ϕ =  η 2 ζ32 η  : C̃a′ → C̃a , η ζ32 η 2 ζ32 and the twist C˜a′ has a non-singular plane model C̃a′ : 2(x5 z + z 5 y + ζ3 y 5 x) − 5(1 + ζ3 )(y 4 z 2 + x2 z 4 ) + 9x4 y 2 + 20ζ3 (x3 yz 2 + x2 y 3 z) − 20(ζ3 + 1)xy 2 z 3 + +a(−(x5 z + z 5 y + ζ3 y 5 x) + 2x4 y 2 − 2(ζ3 + 1)(x2 z 4 + y 4 z 2 ) − ζ3 (x3 yz 2 − x2 y 3 z) + (ζ3 + 1)xy 2 z 3 ). If q ≡ 1 mod 9, the same ϕ works, but this time the cocycle becomes trivial since η ∈ Fq . 6. Twists of smooth plane curves with diagonal cyclic automorphism group We observed in Remark 4.4, that the algorithm for computing T wistk (C) described in [16] can be substantially improved if the smooth curve C over k admits a non-singular plane model and such that the morphism Σ in theorem 4.1 is trivial. In this section, we apply this algorithm for curves C having an extra property: there exists a plane k-model FC (X, Y, Z) = 0 having a diagonal cyclic automorphism group, i.e., we have that Aut(FC ) =< α > with α a diagonal matrix. In such case, we prove that all the elements of T wistk (FC = 0) are given by non-singular plane models of the form FDC = 0 with D a diagonal matrix. We apply this method to some particular families of smooth plane curves. Definition 6.1. Consider a smooth plane curve C over k with a non-singular plane model over k given by FC (X, Y, Z) = 0. We say that C ′ ∈ T wistk (C) is a diagonal twist of C if there exist M ∈ PGL3 (k) and D a diagonal matrix in PGL3 (k) such that C ′ is k-isomorphic to FM DC (X, Y, Z) = 0. The condition of having cyclic automorphism group is not enough to ensure that all the twists having plane non-singular models, are diagonal twists. We will show an example. 6.1. Diagonal cyclic automorphism group: all twists are diagonal. Motivated by the results in Section 4 and following the philosophy of the third author’s thesis in [15], we prove the next result. ON TWISTS OF SMOOTH PLANE CURVES 13 Theorem 6.2. Let C : FC (X, Y, Z) = 0 be a non-singular plane curve defined over k. Assume that Aut(FC ) ⊆ PGL3 (k) is a non-trivial cyclic group of exact order n (prime with the characteristic of k) generated by an element α = diag(1, ζna , ζnb ) where a, b ∈ N. Then all the twists Twistk (C) are given by plane equations of the form FDC (X, Y, Z) = 0 with FDC (X, Y, Z) ∈ k[X, Y, Z] and D is a diagonal matrix. In particular, the map Σ is trivial. Proof. We just need to notice that the map Σ in Theorem 4.1 factors as follows: Σ : H1 (k, Aut(FP C )) → (H1 (k, GL1 (k)))3 → H1 (k, GL3 (k)) → H1 (k, PGL3 (k)). Hence, Σ is trivial and all the cocycles are given by diagonal matrices.  Remark 6.3. A general statement of the above result is as follows: Assume that C is a curve defined over k with a plane non-singular model over k say, FC (X, Y, Z) = 0, and that there exists [C ′ ] ∈ T wistk (C) such that C ′ admits a non-singular plane model over k given by FQC (X, Y, Z) = 0 with Aut(FQC ) = ⟨diag(1, ξna , ξnb )⟩. Then, any [C ′′ ] ∈ T wistk (C) has a representative C ′′ given by a non-singular plane model over k of the form FQDC = 0 with D diagonal. In particular, Aut(FQDC ) is diagonal, and the twists of C over k are diagonal. 2 We apply now Theorem 6.2 to some particular smooth plane curves with cyclic automorphism group. These twists are also computed in the appendix by the algorithm in [16] with the improvement of Theorem 4.1 in order to run the algorithm in [16] in PGL3 instead of PGLg where g the genus of the smooth plane curve. For a finite group G, we denote by M^ P l (G) the elements, in the moduli space M of smooth, genus g curves g g over k, that admit a non-singular plane model, and their full automorphism group is isomorphic to G. The strata M^P l (G) is the disjoint union of the different components ρ(M g ^P l (G)) where ρ denote different non-conjugate g injective representations of G inside PGL3 (k), we refer to [1] for complete details. Let k be a field of characteristic 0 or p > (d − 1)(d − 2) + 1 (see the last section in [1] and [3]). Then we have: ^− 1)Z) = {X d + Y d + XZ d−1 = 0}, MgP l (Z/d(d ^ MgP l (Z/(d − 1)2 Z) = {X d + Y d−1 Z + XZ d−1 = 0}. Both curves are defined over k and they have a non-singular plane model over k whose automorphism groups are cyclic diagonal of orders d(d − 1) and (d − 1)2 respectively. These groups are generated by d−1 d (d−1)(d−2) diag(1, ζd(d−1) , ζd(d−1) ), and diag(1, ζ(d−1)2 , ζ(d−1)2 ), respectively, see Theorem 1 in [3]. Theorem 6.4. Let k be a field of characteristic zero or p > (d − 1)(d − 2) + 1. Consider the curve C : X d +Y d +XZ d−1 = 0 with d ≥ 5. Then, the set T wistk (C) is parameterized by A1 := (k ∗ \k ∗ )×(k ∗ \k ∗ )/ ∼, d d−1 where (M, N ) ∼ (M ′ , N ′ ) if and only if M ′ = nd M, N ′ = nmd−1 N for some n, m ∈ k. More precisely, a pair 2Here we present a different proof of Theorem 6.2 assuming the general statement: Consider the natural map Σ : H 1 (k, Aut(FP QC )) → H 1 (k, PGL3 (k)), then by Theorem 4.1, any twist that has a non-singular plane model over k is given by some Q′ ∈ PGL3 (k), i.e. it has a non-singular plane model over k of the form FP QQ′ C = 0. This, in particular, defines the cocycle σ 7→ Q′ ·σ (Q′ )−1 , which is trivial in H 1 (k, PGL3 (k)) because Q′ ∈ PGL3 (k). Now, Q′ ·σ Q′−1 ∈ Aut(FP QC ) = ⟨diag(1, ξn a , ξ b )⟩ ( observe that Q′−1 : F n P QC → FP QQ′ C ), thus σ (Q′−1 ) = a′ , ξ b′ ) for some integers Q′−1 diag(1, ξn a′ , b ′ . Writing Q′−1 = (ai,j ), one easily deduces that σ(ai,j ) = uj ai,j with u1 = 1, u2 = n ′ ′ a , u = ξ b ; therefore fixing j, we get ξn 3 n σ(ai,j )a−1 −1 −1 i,j = σ(ai′ ,j )ai′ ,j . In particular, ai,j ai′ ,j is invariant under all σ ∈ Gk , and thus ai,j = qi ai′ ,j for some qi ∈ k. Then, we can consider Q′ = DM with D diagonal over k and M ∈ PGL3 (k). Now, the k-model FP QDM C (X, Y, Z) = 0 is k-isomorphic to FP QDC (X, Y, Z) = 0, which is a diagonal twist of FP QC (X, Y, Z) = 0. Thus, all the twists of C over k admitting a non-singular plane model over k are diagonal. But, also the morphism Aut(FP QC ) =< a , ξ b ) >,→ PGL (k) factors through GL (k), and H 1 (k, GL (k)) is trivial. Hence Σ is the zero map, and all the twists of diag(1, ξn n 3 3 3 C over k admit also a non-singular plane model over k. This completes the proof. 14 E. BADR, F. BARS, AND E. LORENZO (M, N ) ∈ A1 corresponds to a twist of the form M X d + Y d + N XZ d−1 = 0 and vice versa, i.e, every twist corresponds to some pair (M, N ) ∈ A1 . Proof. The description of the twists is an easy consequence of Theorem 4.1. Finally, the equivalence of twists comes from the fact that two such cohomologous cocycles are related by a diagonal matrix.  Similarly, we have: Theorem 6.5. Let k be a field of characteristic zero or positive characteristic > (d − 1)(d − 2) + 1. Consider the curve X d + Y d−1 Z + XZ d−1 = 0 with d ≥ 5. Then, the set T wistk (C) is parameterized by A2 := (k ∗ \ k ∗ ) × d−1 (k ∗ \ k ∗ )/ ∼, where (M, N ) ∼ (M ′ , N ′ ) if and only if M ′ = nd−1 mM, N ′ = md−1 N for some n, m ∈ k. That d−1 is, we associate to an (M, N ) ∈ A2 the twist X d + M Y d−1 Z + N XZ d−1 = 0, and vice versa. Remark 6.6. In the work in [3] (or in [2], for curves of genus 6), the first two authors give different families of curves depending on certain parameters for ρ(MgP^ l (Z/nZ)) with ρ(Z/nZ) generated by a diagonal matrix. So, it is possible to apply Theorem 6.2 to compute the twists of curves in these families. In order to get a precise parametrization of the twists, we need to deal with representative families for such strata ρ(M P^ g l (Z/nZ)) (see the ideas of Lercier, Ritzenthaler, Rovetta and Sisjling in [14]). In the upcomming work [4], we study representative families for the different stratas of ρ(M^P l (G)). 6 6.2. Aut(C) cyclic does not imply diagonal twists. The hypothesis that the automorphism group of C is generated by a diagonal matrix on Theorem 6.2 cannot be removed. Example 6.7. Consider the non-singular plane curve C0 defined over Q by the equation X 4 Y + Y 4 Z + XZ 4 + (X 3 Y 2 + Y 3 Z 2 + X 2 Z 3 ) = 0 and denote such model by FC0 (X, Y, Z) = 0. Lemma 6.8. The group Aut(FC0 ) is isomorphic to Z/3Z (then so does Aut(C0 )), and it is generated by [Y ; Z; X] as a subgroup of PGL3 (Q). Proof. We have s := [Y ; Z; X] ∈ Aut(FC0 ) is of order 3, therefore the group Aut(FC0 ) should be one of the groups in [3, Table 2] with 3 dividing the order. First, we note that there are no elements t ∈ Aut(FC0 ) of order 2 such that tst = s−1 : To show this we consider, for simplicity, the Q-equivalent model FP C0 of the form ( √ √ ) 4X 5 + 20X 3 Y Z + (−5 − 9i 3)Y 3 + (−5 + 9i 3)Z 3 X 2 − 6XY 2 Z 2 − 4Y Z(Y 3 + Z 3 ), through the transformation P of the shape   1 1 1    1 ξ3 ξ32  . 1 ξ32 ξ3 Recall that Aut(FP C0 ) = P −1 Aut(FC0 )P , in particular s′ := P −1 sP = diag(1; ξ3 ; ξ32 ) ∈ Aut(FP C0 ). Now, if t′ ∈ P GL3 (Q) is of order 2 such that t′ s′ t′ = s′−1 , then t′ should be of the shapes [X; aZ; a−1 Y ], [aZ; Y ; a−1 X] or [aY ; a−1 Y ; Z] for some a ∈ Q. But non of these transformations retains the defining equation FP C0 = 0, hence S3 does not occur as a bigger group of automorphisms. Then so are the groups GAP (30, 1) and GAP (150, 5), as both groups contain an S3 and also there exists a single conjugacy class of elements of order 3 inside these groups. In particular, we get the same conclusion for Aut(FC0 ), which was to be shown. Second, assume that Aut(FC0 ) is conjugate, through some P ∈ PGL3 (Q) to GAP (39, 1) such that FP C0 (X, Y, Z) = X Y +Y 4 Z+XZ 4 . Now, any element of order 3 in GAP (39, 1), with respect to the given representation in [2, Ta- 4 ble 2], is conjugate to s or s−1 . Then we may impose, without loss of generality, that P −1 sP = s, since s is not conjugate to s−1 in PGL3 (Q). In particular, P has the shape   α1 α2 α3  r   ζ3 α3 ζ3r λα1 ζ3r α2  ∈ PGL3 (Q), ζ3−r α2 ζ3−r α3 ζ3−r α1 ON TWISTS OF SMOOTH PLANE CURVES 15 When r = 0, we need to cancel the coefficients of Y 5 , Y 4 X, Y 3 X 2 , Y 3 Z 2 , and Y 3 XZ in FP C0 , which is impossible due to invertibility of P . Furthermore, if r = 1 or 2, we also force the coefficients of X 5 , Y 5 and Z 5 in FP C0 to be zeros, and then we obtain a diagonal transformation P . In particular, the defining equation FP C0 = 0 is not the claimed one. That is Aut(FC0 ) can not be conjugate to GAP (39, 1). Hence, Aut(FC0 ) is cyclic of order 3, and we are done.  Lemma 6.9. There exists a non-diagonal twist C0 over Q. Proof. The defining equation FC0 = 0 has degree 5, thus any twist of C0 admits also a non-singular plane model over Q defined by FP C0 (X, Y, Z) = 0 for some P ∈ PGL3 (Q). We construct the twist following the classical algorithm in [16] and Theorem 4.1 because Σ is trivial. Firstly, the twisted product Γ := Aut(FC0 ) o Gal(K/Q) is isomorphic to Z/3Z (recall that the field K, where all automorphisms of FC0 are defined, is Q). The group Aut(FC0 ) is identified with lat[2] (see notation in the code) inside the lattice of subgroups of Γ. Then, if we modify the MAGMA code in [15, Table 5.5] to fit our case, then we deduce that FC0 has exactly two non-trivial twists for each cyclic cubic field extension L/Q. Since the set of such extensions is not empty, the curve C0 has a non-trivial twist. Secondly, a twist of FC0 through a diagonal isomorphism D ∈ PGL3 (Q) is always trivial (then so are all the twists of the form FDM C0 with M ∈ PGL3 (Q)). On the other hand, to obtain a non-trivial twist of the form FM DC0 with D = diag(1; a; b) ∈ PGL3 (Q) and M = (aij ) ∈ PGL3 (Q), we must satisfy the 1-cocycle condition (M D).h (M D)−1 = s (recall that ϕ = (M D)−1 : FC0 → FM DC0 ), where h is a generator of Gal(L/Q). In particular,     a11 aa12 ba13 a21 h(a)a22 h(b)a23      a21 aa22 ba23  = λ  a31 h(a)a32 h(b)a33  a31 aa32 ba33 a11 h(a)a12 h(b)a13 for some λ ∈ Q. From the 1st column, we get λa21 = a11 , λa31 = a21 , and λa11 = a31 . Hence λ3 − 1 = 0, but √ also aij ∈ Q, then λ = 1. Consequently, h(a3 ) = a3 , and h(b3 ) = b3 . That is a3 , b3 ∈ L⟨h⟩ (= Q), thus a = 3 N , √ √ √ and b = 3 N ′ for some N, N ′ ∈ Q. In particular, the twist ϕ has Q( 3 N , 3 N ′ ) as its splitting field, which is not Galois if a or b does not belong to Q, a contradiction. Consequently, C0 has a non-trivial twist, which can not be obtained through any diagonal isomorphism modulo PGL3 (Q).  Therefore we obtain, Proposition 6.10. Let C be a non-singular plane curve over k, a field of characteristic zero, admitting a non-singular plane model FC (X, Y, Z) = 0 over k such that Aut(FC ) ⊆ PGL3 (k) is a cyclic group of order n generated by a matrix α, and no element in the conjugacy class of α in PGL3 (k) is neither a diagonal matrix. Then the twists mapping to zero by Σ (i.e., those ones admitting a plane non-singular model over k), are not necessarily diagonal twist. Remark 6.11. The above example in §6.2 extends to positive characteristic p, for p > (d − 1)(d − 2) + 1 with d = 5, (Lemma 6.8 remains true by the arguments in [1, §6]), and ζ3 ∈ / k in order to construct the non-trivial diagonal twist in the proof of Lemma 6.9. Remark 6.12. Degree 5 is the smallest degree for which such an example exists, see the third author thesis [15] to discard degree 4 exceptions. Appendix A. The classical algorithm on twists for non-hyperelliptic curves The algorithm for computing T wistk (C) of a non-hyperelliptic curve C of genus g ≥ 3 developed in [15, Chp.1] and [16] has three main steps: (1) canonical model of C, (2) Solutions of the Galois embedding problem, (3) Explicit equation of Twists. 16 E. BADR, F. BARS, AND E. LORENZO Assume that C is a smooth curve over k with a plane non-singular model over k such that Σ is trivial, in such case all the twist admits a plane non-singular model over k, see Theorem 4.1. Now, instead of computing a canonical model of C, we can consider a plane model over k associated to C, modifying point (1) of the algorithm. The point (2) is independent of the embedding of C inside a projective space. In point (3), the algorithm of [16] requires to investigate the solutions in GLg (k) using [15, Lemma 1.1.3]. Now, in the modified algorithm, it is enough to look for solutions in GL3 (k). As in [15, 16], the isomorphisms covering the solutions of the Galois embedding problems in GLg (k) or GL3 (k) is quite hard except that we have a control of the matrix shape that could appear. We use the above modified algorithm to compute the twist over k for the curve C : X 5 + Y 5 + XZ 4 = 0 where k is a field of zero characteristic or positive characteristic > (5 − 1)(5 − 2) + 1 = 13. Here we recover the result obtained for such curve in Theorem 6.4. The automorphism group Aut(C) ∼ = GAP (20, 2) is generated by the elements s := diag(1; 1; i), and t := diag(1; ζ5 ; 1), so it is defined over K = k(i, ζ5 ) (we assume the generic case in which i, ζ5 ∈ / k). Then, [K : k] = 8 and the Galois group Gal(K/k) is generated by τ1 : i 7→ −i, ζ5 7→ ζ5 and τ2 : i 7→ i, ζ5 7→ ξ52 of order 2 and 4 respectively with τ2 τ1 = τ1 τ2 (in particular, Gal(K/k) ∼ = GAP (8, 2)). The group Gal(K/k) acts naturally on Aut(C) as: τ1 : s 7→ s3 , t 7→ t and τ2 : s 7→ s, t 7→ t2 . The twisted product Γ := Aut(C)oGal(K/k) is isomorphic to GAP (160, 207), and generated by the elements x := (st, 1), y := (1, τ1 ) and z := (1, τ2 ), where x20 = y 2 = z 4 = 1, yxy = x11 , xz = zx13 and yzy = z. The degree of the defining equation of C is coprime with 3, thus, by Corollary 3.7, every twist of C has a non-singular plane model over k. Consequently, by Theorem 4.1, the map Σ is trivial. In particular, we only look for solutions of the Galois embedding problems, generated by a similar MAGMA code as in [15, Table 5.5], inside GL3 (k), not in GL6 (k). One finds that all the twists of C over k are covered by diagonal matrices, and they are of the form αX d + Y d + βXZ d−1 for some α, β ∈ k through an isomorphism of the shape diag(a; 1; c) in GL3 (k). We thus collect the computations into the following result: Theorem A.1. (Galois embedding problems) The set T wistk (C) is completely determined by table 1. Remark A.2. The Galois embedding problems for C are given by the pairs (G, H) appearing in the second and the third columns in GAP notations. This means that a twist ϕ : C → C ′ of C over k has a splitting field L such that Gal(L/k) ∼ = G and Gal(L/K) ∼ = H for some pair (G, H) in the list. By the aid of Proposition 4.1 in [15], we find solutions to these Galois embedding problems as described in the 8th column, where N ∈ k ∗ \ k ∗2 and M ∈ k ∗ \ k ∗5 . We also provide generators of G and H in the 4th , 5th , and 6th columns: G is generated by the elements h o 1 ∈ H o 1 and gi o τi with i = 1, 2. The integer n(G,H) that appears in the 7th column is the number of non-equivalent twists of C with the same splitting field L. In the remaining part of the table, we give the associated set of non-equivalent twists which are defined by equations of the form αX d + Y d + βXZ d−1 = 0 √ through an isomorphism of the form diag(a; 1; c) with a = 5 α, r ∈ {1, 25}, ℓ ∈ {1, 2, 3, 4}, ji ∈ {0, 1}. ON TWISTS OF SMOOTH PLANE CURVES 17 Table 1. The pairs (G, H), and Twists ID(G) ID(H) gen (H) g1 g2 n(G,H) L α β φ 1 s 1 −4 2 s s2 −100 < 8, 2 > < 1, 1 > 1 1 K 1 2 3 1 s 25 c= 4β 4 1 1 1 √ √ 5 < 16, 10 > s 1 K( N) −4rN 2 6 < 16, 3 > s s √ −20rN 2 4 < 2, 1 > s2 2 K( 5N 2 ) 1 7 < 16, 3 > 1 s 5rN 2 √ 8 < 16, 10 > 1 1 K( N) rN 2 9 1 s2 25M ℓ c = 4 βℓ M ℓ √ 10 1 1 Mℓ √ M 5 < 40, 12 > < 5, 1 > t 4 K( M) Mℓ 11 s s2 −100M ℓ √ 5 12 s 1 −4M ℓ √ 4 13 < 32, 25 > < 4, 1 > s 1 1 8 K( N) 1 (−4)j1 (25)j2 N 2j3 +1 c = 4 β Mℓ √ √ 5 14 < 80, 50 > s 1 K( N , M ) −4rN 2 M ℓ √ √5 15 < 80, 34 > 1 s √ 125rN 2 M ℓ 4 √ 5 < 10, 2 > s2 , t 8 K( 5N 2 , M ) Mℓ 16 < 80, 34 > s s −500rN 2 M ℓ √ √ 5 17 < 80, 50 > 1 1 K( N , M ) rN 2 M ℓ √ 4 √ 5 18 < 160, 207 > < 20, 2 > s, t 1 1 32 K( N, M) Mℓ (−4)j1 (25)j2 N 2j3 +1 M ℓ References [1] Badr E. and Bars F.; On the locus of smooth plane curves with a fixed automorphism group, see an old version in arXiv:1510.06186 (2015). 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Sijsling, J.Voight; On explicit descent of marked curves and map, arXiv:1504.02814v2 (2015). [23] E. Tengan; Central Simple Algebras and the Brauer group , XVII Latin American Algebra Colloquium, (2009). See the book in http://www.icmc.usp.br/ etengan/algebra/arquivos/cft.pdf [24] L.C.Washington; Introduction to cyclotomic fields, GTM 83, Second edition, Springer, (1997). [25] J.H.M.Wedderbum; On division algebras. Trans.Amer.Math.Soc.22(1921). [26] A. Weil, The field of definition of a variety. American J. of Math. vol. 78, n3 (1956), 509–524. • Eslam Essam Ebrahim Farag Badr Departament Matemàtiques, Edif. C, Universitat Autònoma de Barcelona, 08193 Bellaterra, Catalonia E-mail address: eslam@mat.uab.cat Department of Mathematics, Faculty of Science, Cairo University, Giza-Egypt E-mail address: eslam@sci.cu.edu.eg • Francesc Bars Cortina Departament Matemàtiques, Edif. C, Universitat Autònoma de Barcelona, 08193 Bellaterra, Catalonia E-mail address: francesc@mat.uab.cat • Elisa Lorenzo Universiteit Leiden, 2300 Leiden, The Nederlands E-mail address: e.lorenzo.garcia@math.leidenuniv.nl